ssm et chapter 12

8/8/2019 SSM ET Chapter 12

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24

CHAPTER 12

Vector-Valued Functions

EXERCISE SET 12.1

1. (, +); r() = i 3 j 3.[2, +); r(3) = i ln 3j + k

5. r = 3 cos t i + (t + sin t) j 7.x = 3t2

, y = 29. the line in 2-space through the point (3, 0) and parallel to the vector 2i + 5j

11. the line in 3-space through the point (0,3, 1) and parallel to the vector 2i + 3k

13. an ellipse in the plane z = 1, center at (0, 0, 1), major axis of length 6 parallel to y-axis, minoraxis of length 4 parallel to x-axis

15. (a) The line is parallel to the vector 2i + 3j; the slope is 3/2.(b) y = 0 in the xz-plane so 1 2t = 0, t = 1/2 thus x = 2 + 1/2 = 5/2 and z = 3(1/2) = 3/2;

the coordinates are (5/2, 0, 3/2).

17. (a) y

x

(0, 1)

(1, 0)

(b)

x

y

(1, 1)

(1, 1)

19. r = (1 t)(3i + 4j), 0 t 1 21. x = 2

2

x

y

23. (x 1)2 + (y 3)2 = 1

1

3

x

y

25. x2 y2 = 1, x 1

1

2

x

y


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Exercise Set 12.1 25

27.

(0, 2, /2)

(2, 0, 0)

y

x

z 29.

2

y

x

z

31. False. It is the intersection of the domains of the components.

33. True. See equation (8).

35. x = t, y = t, z = 2t2

x

y

z

z=x2+y

2

x!y=0

37. r = ti + t2j +1

3

81 9t2 t4 k

x

y

z

y=x29x2+y2+9z2=81

39. x2 + y2 = (t sin t)2 + (t cos t)2 = t2(sin2 t + cos2 t) = t2 = z

41. x = sin t, y = 2 cos t, z =

3sin t so x2 + y2 + z2 = sin2 t + 4 cos2 t + 3 sin2 t = 4 and z =

3x; itis the curve of intersection of the sphere x2 + y2 + z2 = 4 and the plane z =

3x, which is a circle

with center at (0, 0, 0) and radius 2.

43. The helix makes one turn as t varies from 0 to 2 so z = c(2) = 3, c = 3/(2).

45. x2 + y2 = t2 cos2 t + t2 sin2 t = t2, x2 + y2 = t = z; a conical helix.47. (a) III, since the curve is a subset of the plane y = x

(b) IV, since only x is periodic in t, and y, z increase without bound

(c) II, since all three components are periodic in t

(d) I, since the projection onto the yz-plane is a circle and the curve increases without bound inthe x-direction


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26 Chapter 12

49. (a) Let x = 3 cos t and y = 3 sin t, then z = 9 cos2 t. (b) z

yx

EXERCISE SET 12.2

1. 1/3, 0 3. 2i 3j + 4k

5. (a) continuous, limt0

r(t) = 0 = r(0) (b) not continuous, limt0

(1/t) does not exist

7.

x

y

r'(p/4)

r''(p)

r(2p) r(3p/2)

9. r(t) = sin tj

11. r(t) = 1, 2t,r(2)= 1, 4r(2)= 2, 4

2

4

1, 4

x

y

13. r(t) = sec t tan ti + sec2 tj,

r(0)= j

r(0)= i

1.5

x

y

1

1


4/19


15. r(t) = 2cos ti 2sin tk,r(/2) = 2k,r(/2)= 2i +j

r (6)= 2 k

y

x

z

(2, 1, 0)

17.

1.500

1.5

19. r(t) = 2ti 1tj, r(1) = 2i j, r(1) = i + 2j; x = 1 + 2t, y = 2 t

21. r(t) = 2 sin ti + 2 cos tj + 3k, r(1/3) = 3 i + j + 3k,r(1/3) = i +

3j + k; x = 1 3 t, y = 3 + t, z = 1 + 3t

23. r(t) = 2i +3

2

3t + 4j, t = 0 at P0 so r

(0) = 2i +3

4j,

r(0) = i + 2j; r = (i + 2j) + t

2i +3

4j

25. r(t) = 2ti +1

(t + 1)2j 2tk, t = 2 at P0 so r(2) = 4i +j + 4k,

r(

2) = 4i +j; r = (4i +j) + t(

4i +j + 4k)

27. (a) limt0

(r(t) r(t)) = i j + k(b) lim

t0(r(t) r(t)) = lim

t0( cos ti sin tj + k) = i + k

(c) limt0

(r(t) r(t)) = 0

29. r1 = 2i + 6tj + 3t

2k, r2 = 4t

3k, r1 r2 = t7;

d

dt(r1 r2) = 7t

6 = r1 r2 + r

1

r2

r1 r2 = 3t6i 2t5j, ddt

(r1 r2) = 18t5i 10t4j = r1 r2 + r1 r2

31. 3ti + 2t2j + C 33.

cos t,

sin t

+ C

35.

1

2sin2t,1

2cos2t

/20

= 0, 1

37.

2

0

t2 + t4 dt =

2

0

t(1 + t2)1/2 dt =1

3

1 + t2

3/2 20

= (5

5 1)/3


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28 Chapter 12

39.

2

3t3/2i + 2t1/2j

91

=52

3i + 4j

41. False. The limit only exists if r(t) is differentiable at t = a. As with functions of a single variable,continuity does not imply differentiability. For example, r(t) = |t|, 0 is continuous at t = 0, butnot differentiable there.

43. True. Equations (11) and (12) express

b

a

r(t) dt as a vector, whose components are the definite

integrals of the components of r(t).

45. y(t) =

y(t) dt = t2i + t3j + C, y(0) = C = i j, y(t) = (t2 + 1)i + (t3 1)j

47. y(t) =

y(t) dt = ti + etj + C1, y

(0) = j + C1 = j so C1 = 0 and y(t) = ti + etj.

y(t) =

y(t) dt =

1

2t2i + etj + C2, y(0) = j + C2 = 2i so C2 = 2i j and

y(t) =1

2 t2

+ 2

i + (et 1)j

49. (a) 2t t2 3t = 2, t2 + t 2 = 0, (t + 2)(t 1) = 0 so t = 2, 1. The points of intersectionare (2, 4, 6) and (1, 1,3).

(b) r = i + 2tj 3k; r(2) = i 4j 3k, r(1) = i + 2j 3k, and n = 2i j + k is normal tothe plane. Let be the acute angle, then

for t = 2: cos = |n r|/(n r) = 3/156, 76;for t = 1: cos = |n r|/(n r) = 3/84, 71.

51. r1(1) = r2(2) = i +j + 3k so the graphs intersect at P; r1(t) = 2ti +j + 9t

2k and

r2(t) = i + 12tj k so r1(1) = 2i + j + 9k and r2(2) = i + j k are tangent to the graphs at P,

thus cos =r1(1) r

2(2)

r1

(1) r2

(2) = 6

86

3, = cos1(6/

258) 68.

53.d

dt[r(t) r(t)] = r(t) r(t) + r(t) r(t) = r(t) r(t) + 0 = r(t) r(t)

55. In Exercise 54, write each scalar triple product as a determinant.

57. Let r1(t) = x1(t)i + y1(t)j + z1(t)k and r2(t) = x2(t)i + y2(t)j + z2(t)k, in both (6) and (7); showthat the left and right members of the equalities are the same.

EXERCISE SET 12.3

1. r(t) = 3t2i + (6t 2)j + 2tk; smooth

3. r(t) = (1 t)eti + (2t 2)j sin(t)k; not smooth, r(1) = 0


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5. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = (3cos2 t sin t)2 + (3 sin2 t cos t)2 + 02 = 9 sin2 t cos2 t,

L =

/20

3sin t cos t dt = 3/2

7. r(t) = et,et,2, r(t) = et + et, L =

1

0

(et + et) dt = e e1

9. r(t) = 3t2i +j + 6 tk, r(t) = 3t2 + 1, L =

3

1

(3t2 + 1) dt = 28

11. r(t) = 3sin ti + 3 cos tj + k, r(t) = 10, L =

2

0

10 dt = 2

10

13. (dr/dt)(dt/d) = (i + 2tj)(4) = 4i + 8tj = 4i + 8(4 + 1)j;

r() = (4 + 1)i + (4 + 1)2j, r() = 4i + 2(4)(4 + 1)j

15. (dr/dt)(dt/d) = (eti 4etj)(2) = 2 e2i 8 e2j;r() = e

2

i + 4e2

j, r() = 2 e2

i

4(2) e

2

j

17. False. r(t) is a scalar, soba

r(t) dt is also a scalar.

19. False; r(s) is undefined for the value of s such that r(s) = 0. For example, we may take

r(s) =s2

i +|s|

2j. Then r(0) is undefined, since |s| is not differentiable at s = 0.

21. (a) r(t) = 2, s =t0

2 dt =

2t; r =

s2

i +s2

j, x =s

2, y =

s2

(b) Similar to part (a), x = y = z =s

323. (a) r(t) = 1, 3, 4 when t = 0,

so s =

t0

1 + 4 + 4 du = 3t, x = 1 + s/3, y = 3 2s/3, z = 4 + 2s/3

(b) r

s=25

= 28/3,41/3, 62/3

25. x = 3 + cos t, y = 2 + sin t, (dx/dt)2 + (dy/dt)2 = 1,

s = t

0

du = t so t = s, x = 3 + cos s, y = 2 + sin s for 0 s 2.

27. x = t3/3, y = t2/2, (dx/dt)2 + (dy/dt)2 = t2(t2 + 1),

s =

t0

u(u2 + 1)1/2 du =1

3[(t2 + 1)3/2 1] so t = [(3s + 1)2/3 1]1/2,

x =1

3[(3s + 1)2/3 1]3/2, y = 1

2[(3s + 1)2/3 1] for s 0


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30 Chapter 12

29. x = et cos t, y = et sin t, (dx/dt)2 + (dy/dt)2 = 2e2t, s =

t0

2 eu du =

2(et 1) so

t = ln(s/

2 + 1), x = (s/

2 + 1) cos[ln(s/

2 + 1)], y = (s/

2 + 1) sin[ln(s/

2 + 1)]

for 0 s 2(e/2 1)

31. dx/dt = a sin t, dy/dt = a cos t, dz/dt = c,

s(t0) = L =t00

a2

sin2

t + a2

cos2

t + c2

dt =t00

a2

+ c2

dt = t0

a2

+ c2

33. x = at a sin t, y = a a cos t, (dx/dt)2 + (dy/dt)2 = 4a2 sin2(t/2),

s =

t0

2a sin(u/2) du = 4a[1 cos(t/2)] so cos(t/2) = 1 s/(4a), t = 2 cos1[1 s/(4a)],

cos t = 2 cos2(t/2) 1 = 2[1 s/(4a)]2 1,sin t = 2sin(t/2) cos(t/2) = 2(1 [1 s/(4a)]2)1/2(2[1 s/(4a)]2 1),x = 2a cos1[1 s/(4a)] 2a(1 [1 s/(4a)]2)1/2(2[1 s/(4a)]2 1),

y =s(8a s)

8afor 0 s 8a

35. (a) (dr/dt)2 + r2(d/dt)2 + (dz/dt)2 = 9e4t, L =

ln 2

0

3e2t dt =3

2e2tln 20

=9

2

(b) (dr/dt)2 + r2(d/dt)2 + (dz/dt)2 = 5t2 + t4 = t2(5 + t2),

L =

2

1

t(5 + t2)1/2 dt = 9 2

6

37. (a) (d/dt)2 + 2 sin2 (d/dt)2 + 2(d/dt)2 = 3e2t, L =

2

0

3et dt =

3(1 e2)

(b) (d/dt)2 + 2 sin2 (d/dt)2 + 2(d/dt)2 = 5, L = 5

1

5 dt = 4

5

39. (a) g() = (b) g() = (1 )

41. Represent the helix by x = a cos t, y = a sin t, z = ct with a = 6.25 and c = 10/, so that theradius of the helix is the distance from the axis of the cylinder to the center of the copper cable,and the helix makes one turn in a distance of 20 in. (t = 2). From Exercise 31 the length of the

helix is 2

6.252 + (10/)2 44 in.

43. r(t) = (1/t)i + 2j + 2tk

(a) r(t) =

1/t2 + 4 + 4t2 =

(2t + 1/t)2 = 2t + 1/t

(b)

ds

dt = 2t + 1/t (c)31 (2t + 1/t) dt = 8 + ln 3

45. If r(t) = x(t)i + y(t)j + z(t)k is smooth, then r(t) is continuous and nonzero. Thus the anglebetween r(t) and i, given by cos1(x(t)/r(t)), is a continuous function of t. Similarly, theangles between r(t) and the vectors j and k are continuous functions of t.


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EXERCISE SET 12.4

1. (a)

x

y (b)

x

y

3. From the marginal note, the line is parametrized by normalizing v, but T(t0) = v/v, sor = r(t0) + tv becomes r = r(t0) + sT(t0).

5. r(t) = 2ti +j, r(t) = 4t2 + 1, T(t) = (4t2 + 1)1/2(2ti +j),T(t) = (4t2 + 1)1/2(2i) 4t(4t2 + 1)3/2(2ti +j);T(1) =

25

i +1

5j, T(1) =

2

5

5(i 2j), N(1) = 1

5i 2

5j.

7. r(t) = 5sin ti + 5 cos tj, r(t) = 5, T(t) = sin ti + cos tj, T(t) = cos ti sin tj;

T(/3) =

3

2i +

1

2j, T(/3) =

1

2i

3

2j, N(/3) =

1

2i

3

2j

9. r(t) = 4sin ti + 4 cos tj + k, T(t) = 117

(4sin ti + 4 cos tj + k),

T(t) =117

(4cos ti 4sin tj), T(/2) = 417

i +117

k

T(/2) = 417

j, N(/2) = j

11. r(t) = et[(cos t sin t)i + (cos t + sin t)j + k], T(t) = 13

[(cos t sin t)i + (cos t + sin t)j + k],

T(t) =1

3[(

sin t

cos t)i + (

sin t + cos t)j],

T(0) =1

3i +

13

j +1

3k, T(0) =

13

(i +j), N(0) = 12

i +1

2j

13. r(t) = cos ti sin tj + tk, r(0) = i, r(0) = j, T(0) = i, so the tangent line has the parametrizationx = s, y = 1.

15. T =3

5cos t i 3

5sin tj +

4

5k, N = sin t i cos tj, B = TN = 4

5cos t i 4

5sin tj 3

5k. Check:

r = 3 cos t i 3sin tj + 4 k, r = 3sin t i 3cos tj, r r = 12 cos t i 12sin tj 9 k,r r = 15, (r r)/r r = 4

5cos t i 4

5sin tj 3

5k = B.

17. r(t) = t sin t i + t cos tj, r = |t|.For t > 0, T = sin t i + cos tj, N = cos t i sin tj;for t < 0, T = sin t i cos tj, N = cos t i + sin tj.In either case, B = T N = k.Check: r = t sin t i + t cos tj, r = (sin t + t cos t) i + (cos t t sin t)j, r r = t2 k,r r = t2; for t = 0, (r r)/r r = k = B.


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32 Chapter 12

19. r(/4) =

2

2i +

2

2j + k, T = sin ti +cos tj =

2

2(i +j), N = (cos ti +sin tj) =

2

2(i +j),

B = k; the rectifying, osculating, and normal planes are given (respectively) by x + y =

2,z = 1,x + y = 0.

21. False. For example, if r(t) = t, 0 then T(t) = 1, 0 is parallel to r(t) for all t > 0.

23. True. By Theorem 12.3.4(b),

r(s)

= 1 for all s, so Theorem 12.2.8 implies that r(s) and r(s)

are orthogonal.

EXERCISE SET 12.5

1. 10.5

= 2

3. (a) At x = 0 the curvature of I has a large value, yet the value of II there is zero, so II is not thecurvature of I; hence I is the curvature of II.

(b) I has points of inflection where the curvature is zero, but II is not zero there, and hence isnot the curvature of I; so I is the curvature of II.

5. r(t) = 2ti + 3t2j, r(t) = 2i + 6tj, =r(t) r(t)

r(t)3 =6t2

(4t2 + 9t4)3/2=

6

|t|(4 + 9t2)3/2

7. r(t) = 3e3ti etj, r(t) = 9e3ti + etj, = r(t) r(t)r(t)3 =

12e2t

(9e6t + e2t)3/2

9. r(t) = 4sin ti + 4 cos tj + k, r(t) = 4cos ti 4sin tj, = r(t) r(t)r(t)3 =

4

17

11. r(t) = sinh ti + cosh tj + k, r(t) = cosh ti + sinh tj, =r(t) r(t)

r(t)

3

=1

2 cosh2 t

13. r(t) = 3sin ti + 4 cos tj + k, r(t) = 3cos ti 4sin tj,r(/2) = 3i + k, r(/2) = 4j; = 4i + 12k/ 3i + k3 = 2/5, = 5/2

15. r(t) = et(cos t sin t)i + et(cos t + sin t)j + etk,r(t) = 2et sin ti + 2et cos tj + etk, r(0) = i +j + k,r(0) = 2j + k; = i j + 2k/i +j + k3 =

2/3, = 3/

2

17. r(s) =1

2cos

1 +

s

2

i 1

2sin

1 +

s

2

j +

3

2k, r(s) = 1, so

dT

ds

=

1

4

sin1 + s2 i

1

4

cos1 + s2j, = dT

ds = 1

4

19. True, by Example 1 with a = 2.

21. False. By equation (1), the curvature is r(s), not r(s). (r(s) is always 1, if it exists.)

23. (a) r = xi + yj, r = xi + yj, r r = |xy xy|, = |xy yx|

(x2 + y2)3/2


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(b) Set x = t, y = f(x) = f(t), x = 1, x = 0, y =dy

dx, y =

d2y

dx2, =

|d2y/dx2|(1 + (dy/dx)2)3/2

25. (x) =| sin x|

(1 + cos2 x)3/2, (/2) = 1

27. (x) =ex

(1 + e2x)3/2, (1) =

e1

(1 + e2)3/2

29. x(t) = 2t, y(t) = 3t2, x(t) = 2, y(t) = 6t,x(1/2) = 1, y(1/2) = 3/4, x(1/2) = 2, y(1/2) = 3; = 96/125

31. x(t) = 1, y(t) = 1/t2, x(t) = 0, y(t) = 2/t3x(1) = 1, y(1) = 1, x(1) = 0, y(1) = 2; = 1/2

33. (a) (x) =| cos x|

(1 + sin2 x)3/2,

(x) =(1 + sin2 x)3/2

| cos x|(0) = () = 1.

c

(0) = ( ) = 1c

x

y

r r

(b) (t) =2

(4sin2 t + cos2 t)3/2,

(t) =1

2

(4sin2 t + cos2 t)3/2,

(0) = 1/2, (/2) = 4

1

2

(6) = 4

(0) =1

2

x

y

35.

x

y

1 2 3 4 5

0.4

x

y

1 2 3 4 5

1

y = f(x) = xex y = (x) =|x 2|ex

[1 + (1 x)2e2x]3/2

37. (a) =|12x2 4|

[1 + (4x3 4x)2]3/2 (b)

f(x)

k

2 2

8

x

y


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34 Chapter 12

(c) f(x) = 4x3 4x = 0 at x = 0,1, f(x) = 12x2 4, so extrema at x = 0,1, and = 1/4for x = 0 and = 1/8 when x = 1.

39. r() =

r sin + cos dr

d

i +

r cos + sin

dr

d

j;

r() =r cos 2sin drd + cos d

2r

d2

i +r sin + 2 cos drd + sin d

2r

d2

j;

=

r2 + 2

dr

d

2 r d

2r

d2

r2 +

dr

d

23/2 .

41. () =3

2

2(1 + cos )1/2, (/2) =

3

2

243. () =

10 + 8cos2 3

(1 + 8 cos2 )3/2, (0) =

2

3

45. Let y = t, then x =t2

4pand (t) =

1/|2p|[t2/(4p2) + 1]3/2

;

t = 0 when (x, y) = (0, 0) so (0) = 1/|2p|, = 2|p|.

47. Let x = 3 cos t, y = 2 sin t for 0 t < 2, (t) = 6(9sin2 t + 4 cos2 t)3/2

so

(t) =1

6(9sin2 t + 4 cos2 t)3/2 =

1

6(5sin2 t + 4)3/2 which, by inspection, is minimum when

t = 0 or . The radius of curvature is minimum at (3, 0) and (3, 0).

49. From Exercise 39: dr/d = aea = ar, d2r/d2 = a2ea = a2r; = 1/[

1 + a2 r].

51. (a) d2y/dx2 = 2, () = |2cos3 |(b) dy/dx = tan = 1, = /4, (/4) = |2cos3(/4)| = 1/2, = 2(c)

2 1

3

x

y

53. = 0 along y = 0; along y = x2, (x) = 2/(1 + 4x2)3/2, (0) = 2. Along y = x3,

(x) = 6|x|/(1 + 9x4)3/2, (0) = 0.

55. = 1/r along the circle; along y = ax2, (x) = 2a/(1 + 4a2x2)3/2, (0) = 2a so 2a = 1/r,a = 1/(2r).


12/19


57. Let y(x) =

f(x) if x 0;ax2 + bx + c if x > 0.

Since (x) =|y|

(1 + y2)3/2, the transition will be smooth ify, y, and y are all continuous at x = 0.

This happens if f(0) = c, f(0) = b, and f(0) = 2a. So if we let a = f(0)/2, b = f(0), andc = f(0), the transition will be smooth.(Note that we dont need f(x) to exist for all x 0; it suffices to have f(x) continuous.)

59. (a) B dB

ds= 0 because B(s) = 1 so dB

dsis perpendicular to B(s).

(b) B(s) T(s) = 0, so 0 = B(s) dT

ds+

dB

ds T(s). Since

dT

ds= N(s), B(s)

dT

ds=

B(s) N(s) = 0. HencedB

ds T(s) = 0; thus

dB

dsis perpendicular to T(s).

(c)dB

dsis perpendicular to both B(s) and T(s) but so is N(s), thus

dB

dsis parallel to N(s) and

hence a scalar multiple of N(s).

(d) If C lies in a plane, then T(s) and N(s) also lie in the plane; B(s) = T(s) N(s) so B(s)

is always perpendicular to the plane. Since

dB

ds exists at each point on the curve, B is

continuous. Since B = 1, either B = k for all s or B = k for all s; in either case dBds

= 0.

61. r(s) = dT/ds = N so r(s) = dN/ds + (d/ds)N but dN/ds = T + B sor(s) = 2T + (d/ds)N + B, r(s) r(s) = T (N) = T N = B,[r(s) r(s)] r(s) = 3B T + (d/ds)B N + 2B B = 2, = [r(s) r(s)] r(s)/2 = [r(s) r(s)] r(s)/r(s)2 andB = T N = [r(s) r(s)]/r(s)

63. r = 2i + 2tj + t2k, r = 2j + 2tk, r = 2k, r r = 2t2i 4tj + 4k, r r = 2(t2 + 2), = 8/[2(t2 + 2)]2 = 2/(t2 + 2)2

65. r = eti etj +2k, r = eti + etj, r = eti etj, r r = 2eti + 2etj + 2k,

r r = 2(et + et), = (22)/[2(et + et)2] = 2/(et + et)2

EXERCISE SET 12.6

1. v(t) = 3sin ti + 3 cos tja(t) = 3cos ti 3sin tjv(t) =

9sin2 t + 9 cos2 t = 3

r(/3) = (3/2)i + (3

3/2)j

v(/3) = (33/2)i + (3/2)ja(/3) = (3/2)i (33/2)j

3

3

2

3 32( ),

3

2

3 32

a = i j

x

y3 3

2v = i + j

3

2


13/19

36 Chapter 12

3. v(t) = eti etja(t) = eti + etj

v(t) = e2t + e2tr(0) = i +j

v(0) = i ja(0) = i +j

(1, 1)

v = i j

a = i + j

x

y

5. v = i + tj + t2k, a = j + 2tk; at t = 1, v = i +j + k, v = 3, a = j + 2k

7. v = 2sin ti + 2 cos tj + k, a = 2cos ti 2sin tj;at t = /4, v = 2i + 2j + k, v = 5, a = 2i 2j

9. (a) v = a sin ti + b cos tj, a = a2 cos ti b2 sin tj = 2r(b) From part (a), a = 2r

11. If a = 0 then x(t) = y(t) = z(t) = 0, so x(t) = x1t + x0, y(t) = y1t + y0, z(t) = z1t + z0, themotion is along a straight line and has constant speed.

13. v = (6/

t)i + (3/2)t1/2j, v =

36/t + 9t/4, dv/dt = (36/t2 + 9/4)/(2

36/t + 9t/4) = 0

if t = 4 which yields a minimum by the first derivative test. The minimum speed is 3

2 whenr = 24i + 8j.

15. (a)

t

speed

3

6

2!/3

(b) v = 3cos 3t i + 6sin 3tj, v =

9cos2 3t + 36 sin2 3t = 3

1 + 3 sin2 3t; by inspection, max-imum speed is 6 and minimum speed is 3

(d) By inspection, v = 3

1 + 3sin2 3t is maximal when sin 3t = 1; this occurs first whent = /6.

17. v(t) = sin ti + cos tj + C1, v(0) = j + C1 = i, C1 = i j, v(t) = (1 sin t)i + (cos t 1)j;r(t) = (t + cos t)i + (sin t t)j + C2, r(0) = i + C2 = j,C2 = i +j so r(t) = (t + cos t 1)i + (sin t t + 1)j

19. v(t) =

cos ti + sin tj + etk + C1, v(0) =

i + k + C1 = k so

C1 = i, v(t) = (1 cos t)i + sin tj + etk; r(t) = (t sin t)i cos tj + etk + C2,r(0) = j + k + C2 = i + k so C2 = i +j, r(t) = (t sin t 1)i + (1 cos t)j + etk.

21. v = 3t2i + 2tj, a = 6ti + 2j; v = 3i + 2j and a = 6i + 2j when t = 1 so

cos = (v a)/(v a) = 11/130, 15.

23. (a) displacement = r1 r0 = 0.7i + 2.7j 3.4k


14/19


(b) r = r1 r0, so r0 = r1 r = 0.7i 2.9j + 4.8k.

25. r = r(3) r(1) = 8i + (26/3)j; v = 2ti + t2j, s =

3

1

t

4 + t2 dt =13

13 553

.

27. r = r(ln3) r(0) = 2i (2/3)j + 2(ln 3)k; v = eti etj + 2 k, s =

ln 3

0

(et + et) dt =8

3.

29. In both cases, the equation of the path in rectangular coordinates is x2 + y2 = 4, the particles

move counterclockwise around this circle; v1 = 6sin3ti + 6cos 3tj andv2 = 4t sin(t2)i + 4t cos(t2)j so v1 = 6 and v2 = 4t.

31. (a) v = eti + etj, a = eti + etj; when t = 0, v = i + j, a = i + j, v = 2, v a = 0,v a = 2k so aT = 0, aN =

2.

(b) aTT = 0, aNN = a aTT = i + j (c) = 1/

2

33. (a) v = (3t2 2)i + 2tj, a = 6ti + 2j; when t = 1, v = i + 2j, a = 6i + 2j, v = 5, v a = 10,v a = 10k so aT = 2

5, aN = 2

5

(b) aTT = 255

(i + 2j) = 2i + 4j, aNN = a aTT = 4i 2j

(c) = 2/

5

35. (a) v = eti 2e2tj + k, a = eti + 4e2tj; when t = 0, v = i 2j + k, a = i + 4j, v = 6,v a = 7, v a = 4i +j + 6k so aT = 7/

6, aN =

53/6

(b) aTT = 76

(i 2j + k), aNN = a aTT = 136

i +19

3j +

7

6k

(c) =

53

6

6

37. v = 4, v a = 12, v a = 8k so aT = 3, aN = 2, T = j, N = (a aTT)/aN = i

39. aT =d2s

dt2=

d

dt

t2 + e3t =

2t 3e3t2

t2 + e3tso when t = 0, aT = 3

2.

41. aN = (ds/dt)2 = (1/)(ds/dt)2 = (1/1)(2.9 105)2 = 8.41 1010 km/s2

43. aN = (ds/dt)2 = [2/(1 + 4x2)3/2](3)2 = 18/(1 + 4x2)3/2

45. a = aTT + aNN; by the Pythagorean Theorem aN =a2 a2T =

9 9 = 0

47. True. By equation (1) the velocity vector is ds/dt multiplied by the unit tangent vector.

49. False. Equations (10) and (11) imply that a and v are parallel, but they may point in oppositedirections. For example, ifr(t) = (ln t)i +(sin(ln t))j then (1) = 0, v(1) = i +j, and a(1) = ij.

51. From (14),a2 = a a = (aTT+aNN) (aTT+aNN) = a2T(T T)+ 2aTaN(T N)+ a2N(N N) = a2T+a2N,since T and N are orthogonal unit vectors. Hence a2N = a2 a2T. Since aN 0 (from equation(13)), aN =

a2 a2T.


15/19

38 Chapter 12

53. 10 km/h is the same as100

36m/s, so F = 500 1

15

100

36

2 257.20 N.

55. v0 = 80, = 60, s0 = 168 so x = 40t, y = 168 40

3 t 16t2; y = 0 whent = 73/2 (invalid) or t = 3 so x(3) = 403 ft.

57. = 30, s0 = 0 so x =

3v0t/2, y = v0t/2 16t2; dy/dt = v0/2 32t, dy/dt = 0 when t = v0/64

so ymax = v2

0/256 = 2500, v0 = 800 ft/s.

59. v0 = 800, s0 = 0 so x = (800cos )t, y = (800 sin )t 16t2 = 16t(50sin t); y = 0 when t = 0or 50sin so xmax = 40, 000 sin cos = 20, 000sin2 = 10, 000, 2 = 30

or 150, = 15

or 75.

61. (a) From (26), r(t) = (40cos60)t i +

4 + (40 sin 60)t 1

2gt2

j = 20t i + ( 4 + 2 0

3t 16t2)j.

When x = 15, t =3

4, and y = 4 + 20

3 3

4 16

3

4

2 20.98 ft, so the water clears the

corner point A with 0.98 ft to spare.

(b) y = 20 when 16t2

20

3t + 16 = 0, t =5

3 118

, t

0.668 (reject) or 1.497, x(1.497)

29.942 ft, so the water hits the roof.(c) about 29.942 15 = 14.942 ft

63. (a) x = (35

2/2)t, y = (35

2/2)t 4.9t2, from Exercise 23(a) in Section 12.5

=|xy xy|

[(x)2 + (y)2]3/2, (0) =

9.8

352

2= 0.004

2 0.00565685; = 1/ 176.78 m

(b) y(t) = 0 when t =25

14

2, y =

125

4m

65. s0 = 0 so x = (v0 cos )t, y = (v0 sin )t gt2/2(a) dy/dt = v0 sin

gt so dy/dt = 0 when t = (v0 sin )/g, ymax = (v0 sin )

2/(2g)

(b) y = 0 when t = 0 or (2v0 sin )/g, so x = R = (2v20 sin cos )/g = (v

20 sin2)/g when

t = (2v0 sin )/g; R is maximum when 2 = 90, = 45, and the maximum value of R

is v20/g.

67. v0 = 80, = 30, s0 = 5 so x = 40

3t, y = 5 + 40t 16t2

(a) y = 0 when t = (40

(40)2 4(16)(5))/(32) = (530)/4, reject (530)/4 to gett = (5 +

30)/4 2.62 s.

(b) x 403(2.62) 181.5ft.69. (a) v0(cos )(2.9) = 259 cos 23 so v0 cos 82.21061, v0(sin )(2.9) 16(2.9)2 = 259sin23

so v0 sin 11.50367; divide v0 sin by v0 cos to get tan 0.139929, thus 8and v0 82.21061/ cos8 83 ft/s.

(b) From part (a), x 82.21061t and y 11.50367t 16t2 for 0 t 2.9; the distance traveled

is

2.9

0

(dx/dt)2 + (dy/dt)2 dt 268.76 ft.


16/19


EXERCISE SET 12.7

1. (a) From (15) and (6), at t = 0,

C = v0 b0 GMu = v0j r0v0k GMu = r0v20i GMi = (r0v20 GM)i(b) From (22), r0v

20 GM = GMe, so from (7) and (17), v b = GM(cos i + sin j) + GMei,

and the result follows.

(c) From (10) it follows that b is perpendicular to v, and the result follows.

(d) From part (c) and (10), v b = vb = vr0v0. From part (b),v b = GM

(e + cos )2 + sin2 = GM

e2 + 2e cos + 1. By (10) and

part (c), v b = vb = v(r0v0) thus v = GMr0v0

e2 + 2e cos + 1. From (22),

r0v20/(GM) = 1 + e, GM/(r0v0) = v0/(1 + e) so v =

v01 + e

e2 + 2e cos + 1.

(e) From (20) r =k

1 + e cos , so the minimum value of r occurs when = 0 and the maximum

value when = . From part (d) v =v0

1 + e

e2 + 2e cos + 1 so the minimum value of v

occurs when = and the maximum when = 0

3. vmax occurs when = 0 so vmax = v0; vmin occurs when = sovmin =

v01 + e

e2 2e + 1 = vmax 1 e

1 + e, thus vmax = vmin

1 + e

1 e .

5. (a) The results follow from formulae (1) and (7) of Section 10.6.

(b) rmin and rmax are extremes and occur at the same time as the extrema of r2, and henceat critical points ofr2. Thus d

dtr2 = d

dt(r r) = 2r r = 0, and hence r and v = r are

orthogonal.

(c) vmin and vmax are extremes and occur at the same time as the extrema of v2, and henceat critical points of v2. Thus d

dtv2 = d

dt(v v) = 2v v = 0, and hence v and a = v

are orthogonal. By (5), a is a scalar multiple of r and thus v and r are orthogonal.

(d) From equation (2), r v = b and thus b = r v = r v sin . When either r or vhas an extremum, however, the angle = /2 and thus b = rv. Finally, since b isa constant vector, the maximum of r occurs at the minimum of v and vice versa, and thusb = rmaxvmin = rminvmax.

7. r0 = 6440 + 200 = 6640 km so v =

3.99 105/6640 7.75 km/s.

9. From (23) with r0 = 6440 + 300 = 6740 km, vesc =

2(3.99) 105

6740 10.88 km/s.

11. (a) At perigee, r = rmin = a(1 e) = 238,900 (1 0.055) 225,760 mi; at apogee,r = rmax = a(1 + e) = 238,900(1 + 0.055) 252,040 mi. Subtract the sum

of the radius of the Moon and the radius of the Earth to getminimum distance = 225,760 5080 = 220,680 mi,and maximum distance = 252,040 5080 = 246,960 mi.

(b) T = 2

a3/(GM) = 2

(238,900)3/(1.24 1012) 659 hr 27.5 days.

13. (a) r0 = 4000 + 180 = 4180 mi, v =

GM

r0=

1.24 1012/4180 17,224 mi/h


17/19

40 Chapter 12

(b) r0 = 4180 mi, v0 =

GM

r0+ 600; e =

r0v20

GM 1 = 1200

r0

GM+ (600)2

r0GM

0.071;rmax = 4180(1 + 0.071)/(1 0.071) 4819 mi; the apogee altitude is about4819 4000 = 819 mi.

REVIEW EXERCISES, CHAPTER 12

3. the circle of radius 3 in the xy-plane, with center at the origin

5. a parabola in the plane x = 2, vertex at (2, 0,1), opening upward

7. Let r = xi + yj + zk, then x2 + z2 = t2(sin2 t + cos2 t) = t2 = y2

xy

z

11. r(t) = (1 2sin2t)i (2t + 1)j + cos tk, r(0) = i j + k and r(0) = i, so the line is given byx = 1 + t, y = t, z = t.

13. (sin t)i (cos t)j + C

15. y(t) =

y(t) dt = 13 t3i + t2j + C, y(0) = C = i +j, y(t) = (13 t3 + 1)i + (t2 + 1)j

17.

ds

dt

2=

2e2t2

+

2e2t2

+ 4 = 8 cosh2(

2t),

L =

2ln2

0

2

2 cosh(

2t) dt = 2 sinh(

2t)

2ln20

= 2sinh(2ln2) =15

4

19. r = r0 + tP Q= (t 1)i + (4 2t)j + (3 + 2t)k;

drdt = 3, r(s) = s 33 i + 12 2s3 j + 9 + 2s3 k

25. r(t) = 2sin ti + 3 cos tj k, r(/2) = 2i kr(t) = 2cos ti 3sin tj, r(/2) = 3j,r(/2) r(/2) = 3i + 6k and hence, by Theorem 12.5.2b,(/2) =

45/53/2 = 3/5.

27. By Exercise 23(b) of Section 12.5, = |d2y/dx2|/[1 + (dy/dx)2]3/2, but d2y/dx2 = cos x = 0 atx = /2, so = 0.

29. (a) speed (b) distance traveled (c) distance of the particle from the origin


18/19

Making Connections, Chapter 12 41

31. (a) r(t) = 1, so, by Theorem 12.2.8, r(t) is always perpendicular to the vector r(t). Thenv(t) = R( sin ti + cos tj), v = v(t) = R

(b) a = R2(cos ti + sin tj), a = a = R2, and a = 2r is directed toward the origin.

(c) The smallest positive value of t for which r(t) = r(0) satisfies t = 2, so T = t =2

.

33. (a)

dv

dt = 2t

2

i +j + cos 2tk, v0

= i + 2j k, so x(t) =2

3 t

3

+ 1, y(t) = t + 2, z(t) =

1

2 sin2t 1,

x(t) =1

6t4 + t, y(t) =

1

2t2 + 2t, z(t) = 1

4cos2t t + 1

4, since r(0) = 0. Hence

r(t) =

1

6t4 + t

i +

1

2t2 + 2t

j

1

4cos2t + t 1

4

k

(b)ds

dt

t=1

= r(t)t=1

=

(5/3)2 + 9 + (1 (sin 2)/2)2 3.475

35. From Table 12.7.1, GM 3.99 105 km3/s2, and r0 = 6440 + 600 = 7040 km, so

vesc

=2GM

r0 2

3.99

105

7040 10.65 km/s.37. By equation (26) of Section 12.6, r(t) = (60 cos )ti + ((60 sin )t 16t2 + 4)j, and the maximum

height of the baseball occurs when y(t) = 0, 60sin = 32t, t =15

8sin , so the ball clears the

ceiling if ymax = (60sin )15

8sin 16 15

2

82sin2 + 4 25, 15

2 sin2

4 21, sin2 28

75. The ball

hits the wall when x = 60, t = sec , and y(sec ) = 60 sin sec 16sec2 + 4. Maximize theheight h() = y(sec ) = 60 tan 16 sec2 + 4, subject to the constraint sin2 28

75. Then

h() = 60sec2 32 sec2 tan = 0, tan = 6032

=15

8, so sin =

1582 + 152

=15

17, but for

this value of the constraint is not satisfied (the ball hits the ceiling). Hence the maximumvalue of h occurs at one of the endpoints of the -interval on which the ball clears the ceiling, i.e.

0, sin1

28/75

. Since h(0) = 60, it follows that h is increasing throughout the interval, since

h > 0 inside the interval. Thus hmax occurs when sin2 =

28

75, hmax = 60 tan 16sec2 + 4 =

60

2847

16 7547

+ 4 =120

329 1012

47 24.78 ft. Note: the possibility that the baseball keeps

climbing until it hits the wall can be rejected as follows: if so, then y(t) = 0 after the ball hits

the wall, i.e. t =15

8sin occurs after t = sec , hence

15

8sin sec , 15sin cos 8,

15sin2 16, impossible.

MAKING CONNECTIONS, CHAPTER 12

1. (a) The given formulas imply that N(t) = B(t) T(t) = r(t) r(t)

r(t) r(t) r(t)

r(t) .

(b) Since r is perpendicular to r r, Theorem 11.4.5a implies that(r(t) r(t)) r(t) = r(t) r(t)r(t), and the result follows.


19/19

42 Chapter 12

(c) (i) r(t) = 2ti +j, r(1) = 2i +j, r(t) = 2i, u = 2i 4j, N = 15

i 25

j

(ii) r(t) = 4sin t i + 4 cos tj + k, r( 2

) = 4i + k, r(t) = 4cos t i 4sin tj,r(

2) = 4j, u = 17(4j), N = j

3. (a) r(t) = t

0

cosu2

2 du i + t

0

sinu2

2 duj;drdt2

= x(t)2 + y(t)2 = cos2

t2

2

+ sin2

t2

2

= 1 and r(0) = 0

(b) r(s) = cos

s2

2

i + sin

s2

2

j, r(s) = s sin

s2

2

i + s cos

s2

2

j,

(s) = r(s) = |s|(c) (s) +, so the spiral winds ever tighter.

5. r =

a coss

w

i +

a sin

s

w

j +

cs

wk, r =

aw

sins

w

i +

aw

coss

w

j +

c

wk,

r =

a

w2

coss

w i

a

w2

sins

wj, r =

a

w3

sins

w i

a

w3

coss

wj,

r r = ac

w3sin

s

w

i

acw3

coss

w

j +

a2

w3k, (r r) r = a

2c

w6,

r(s) = aw2

, so =c

w2and B =

cw

sins

w

i

cw

coss

w

j +

a

wk

ssm et chapter 12

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